I'm sorry I don't understand you. This sounds like a homework question.

The rules of PF are that you must show effort in working out the solution.

I am saying that I won't give you an answer without seeing at least some work on your part, because I think you are going to copy my solution and use it as your own. Nothing illogical about my reasoning there.

From the global guidelines:

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

(I've highlighted 'textbook style exercises' because this is in the form of a question asked in a textbook.)

Bottom of the sea. An air bubble comes off the ground and becomes a hemisphere. It raises or not? Obviously yes. But a metal-coated hollow hemisphere with almost the density of the air (including the metal coat), it raises or not?

Before you answer, take in mind that they (all?) do not subtract buoyancy at such body shapes, and particular case, "because the water cannot push them upwards" or "because there is no below and above water pressure difference" etc, which seems true if the metal hemisphere does not rise.

Correct me if you are sure enough I am wrong in any of the above thoughts.

Why would the metal sphere not rise if it's density is less than that of the water?

If the object neither rises or falls, it is neutrally buoyant. Which means it's mass = mass it displaces. If you have a metal sphere whose overall density is that of air, it is the equivalent of having a bubble of air, so it will rise (float).

The shape of a submerged body is irrelevant, as long as it displaces a mass of water = to that of it's own mass, it will rise (float). If it does not, it will sink. If the amount of mass displaced = mass of object, it is neutrally buoyant and will neither rise nor sink.

If you read the hydrostatics interperetation of buoyancy, you will see that they imply that the metal hollow hemisphere will not rise (when in contact with the bottom of the sea). Since the air bubble does rise, thιs theory (that says that none of them will rise) is wrong (?). But also the opposite theory (that says that both will rise) cannot be correct if the metal hollow hemisphere does not rise (?). So, my question now is not what physics are saying, but what experiments show: Does the metal hollow hemisphere rise or not?

The questionmarks are because there might be some conditions that the theories do not imply, e.g. the (seeming) theory that says that only the medium-body densities define the phenomenon and not the pressure underneath says that only when the whole body is surrounded by the same medium. So the theoretical answer is not ... known. Please tell me the experiments.

Does the hydrostatics interpretation of buoyancy say this? Or are you simply assuming it says this?

If you want to do a simple experiment, you know a helium balloon will rise, so fill a balloon with helium and sit it on the floor. If it doesn't rise then your assumption is correct, if it does rise (I think you'll find it will), then your assumption is wrong.

As long as the mass of the metal sphere plus contents = mass of air bubble and as such the metal spheres overall density = density of air bubble, then it will rise.

I refer you to this link:
http://www.anzcp.org/CCP/Physics&Chem/Hydrostatic%20Pressure%20&%20Buoyancy%20force.htm [Broken]

If the force F is greater than the downwards force (weight of the object), then the object will rise.

MY assumption is wrong? What assumption? I said that the bubble RISES, and I asked whether the metal-coated hemipshere does rise too, or not. Again, YOU are making assumptions that they both rise.

Your link says:The Buoyant Force [the force diving an object (eg air bubble) upward]:
F = AhD
Force [F] is equal to the area [A] times the height [h] (or depth) times the density of the fluid

Now...is this what...the theory says? The weight of the displaced fluid is AhD, where h is the height...of the body and not of the depth.

What I have grasped from the (somewhere implied) hydrostatics equation with buoyancy, is that Buoyant Force is the (upward force of the fluid towards the down surface of the body)-(downward force of the fluid towards the up surface of the body above)=AhD, where each of these fluid forces is (pressure)A.

MY assumption is wrong? What assumption? I said that the bubble RISES, and I asked whether the metal-coated hemipshere does rise too, or not. Again, YOU are making assumptions that they both rise.

Your link says:
The Buoyant Force [the force diving an object (eg air bubble) upward]:
F = AhD
Force [F] is equal to the area [A] times the height [h] (or depth) times the density of the fluid

Now...is this what...the theory says? The weight of the displaced fluid is AhD, where h is the height...of the body!

What I have grasped from the (somewhere implied) theory, is that Buoyant Force is the (upward force of the fluid towards the down surface of the body)-(downward force of the fluid towards the up surface of the body above)=AhD

I'm slightly concerned you linked me to an article on nuclear weapons. What is wrong with the wikipedia article I link you to. The wiki article explains buoyancy a lot more clearly than this 'history of buoyancy' in your nuclear dossier.

MY assumption is wrong? What assumption? I said that the bubble RISES, and I asked whether the metal-coated hemipshere does rise too, or not. Again, YOU are making assumptions that they both rise.

There's no assumption on my part. I told you very clearly. If the metal sphere has equal mass and as such overall density of the air bubble, it will react identically to the air bubble. Period.

Your link says:
The Buoyant Force [the force diving an object (eg air bubble) upward]:
F = AhD
Force [F] is equal to the area [A] times the height [h] (or depth) times the density of the fluid

Now...is this what...the thory says? The weight of the displaced fluid is AhD, where h is the height...of the body!

That makes no sense at all. Their calcs omit gravity for the reason outlined below. The weight of the displaced fluid = volume of fluid displaced x density x gravity. This must equal the weight of the object = volume of object x density x gravity.

However, as gravity is a constant, you can remove this and compare the mass of the displaced fluid with the mass of the object. If you want to quote forces (weight and buoyancy) gravity must be included.

That nuclear article was saying that the ballon must be pushed underneath. Fortetabouti, check this one, it's better:http://physics.valpo.edu/courses/p111/lectures/lecture21.pdf
(at the chapter 15)
As you see, he did not subtract buoyancy regaring the air that the body displaced above the sea level. And ... it is NOT a simplification as the results are almost the same (I suggest have some doubts if you think that he meant this simplification). So...WHY do you think he did not subtract it?

It is a simplification. You clearly didn't read the wiki article I linked. Here is a quote directly from it:

"Air's density is very small compared to most solids and liquids. For this reason, the weight of an object in air is approximately the same as its true weight in a vacuum. The buoyancy of air is neglected for most objects during a measurement in air because the error is usually insignificant (typically less than 0.1% except for objects of very low average density such as a balloon or light foam)."

The force exerted by the atmosphere on a body such as your cork in the slides, is negligible when compared to the others forces involved.